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There is a definition in my notes and says,

When functions are polynomially bounded, the initial conditions (the value on small inputs) do not make a difference for the solution in asymptotic notations. The initial conditions can make a difference, when the function is not polynomially bounded.

Can somebody explain what that means? Thank you

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2 Answers 2

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If a function $f$ is polynomially bounded it means there exists polynomials $g$ and $h$ such that for all $x$, $$g(x)\le f(x)\le h(x).$$

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  • $\begingroup$ Isn't there always a function h(x) that is greater than f(x)? Can you give an example of a polynomially unbounded function? $\endgroup$ Commented Mar 20, 2013 at 23:32
  • $\begingroup$ E.g. $e^x$. Remember the inequality has to hold for all $x$. $\endgroup$ Commented Mar 20, 2013 at 23:34
  • $\begingroup$ @bigO: $x\mapsto 2^x$ is not polynomially bounded. $\endgroup$ Commented Mar 20, 2013 at 23:34
  • $\begingroup$ @HenningMakholm forr example, if we choose h(x)=3^x, doesn't the inequality hold? I couldnt get the point $\endgroup$ Commented Mar 20, 2013 at 23:37
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    $\begingroup$ @bigO: $3^x$ is not a polynomial! $\endgroup$ Commented Mar 20, 2013 at 23:37
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It depends on the context. In general, $f$ is said polinomially bounded if $|f(x)|\leqslant p(x)$ for some real polynomial $p$. But for example, in the context of the theory of distributions, $f$ is said polinomially bounded if it is smooth and there are real polynomials $p_m$ such that $|d^mf/dx^m|\leqslant p_m(x)$, $m=0,1,2,\dotsc$.

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  • $\begingroup$ The Question asks about functions, so expanding(?) the context to distributions does not seem relevant. In any case the problem is concerned with why initial conditions might matter for asymptotic behavior when functions are not polynomially bounded (but not matter for functions that are). You are welcome to add a new answer if there is something to say about that not covered by the existing Answers (to a nine year old Question). $\endgroup$ Commented Jun 10, 2022 at 5:14

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